Problem: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $q \neq 0$. $p = \dfrac{7q + 14}{q^2 - 7q - 18} \div \dfrac{2q + 20}{q^2 - 9q} $
Explanation: Dividing by an expression is the same as multiplying by its inverse. $p = \dfrac{7q + 14}{q^2 - 7q - 18} \times \dfrac{q^2 - 9q}{2q + 20} $ First factor the quadratic. $p = \dfrac{7q + 14}{(q - 9)(q + 2)} \times \dfrac{q^2 - 9q}{2q + 20} $ Then factor out any other terms. $p = \dfrac{7(q + 2)}{(q - 9)(q + 2)} \times \dfrac{q(q - 9)}{2(q + 10)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ 7(q + 2) \times q(q - 9) } { (q - 9)(q + 2) \times 2(q + 10) } $ $p = \dfrac{ 7q(q + 2)(q - 9)}{ 2(q - 9)(q + 2)(q + 10)} $ Notice that $(q + 2)$ and $(q - 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ 7q(q + 2)\cancel{(q - 9)}}{ 2\cancel{(q - 9)}(q + 2)(q + 10)} $ We are dividing by $q - 9$ , so $q - 9 \neq 0$ Therefore, $q \neq 9$ $p = \dfrac{ 7q\cancel{(q + 2)}\cancel{(q - 9)}}{ 2\cancel{(q - 9)}\cancel{(q + 2)}(q + 10)} $ We are dividing by $q + 2$ , so $q + 2 \neq 0$ Therefore, $q \neq -2$ $p = \dfrac{7q}{2(q + 10)} ; \space q \neq 9 ; \space q \neq -2 $